Enhanced EOC: Problem 35.55 A series RLC circuit consists of a 47.0 ohm resistor

Question

Enhanced EOC: Problem 35.55

A series RLC circuit consists of a 47.0 ohm resistor, a 3.00 mH inductor, and a 510 nF capacitor. It is connected to a 3.0 kHz oscillator with a peak voltage of 4.70 V. You may want to review (pages 1042- 1046) . For help with math skills, you may want to review: Conversion from Degrees to Radians Calculating Trigonometric Function Values What is the instantaneous emf Epsilon when i = I? Express your answer with the appropriate units. Part B What is the instantaneous emf Epsilon when i = 0 A and is decrea Express your answer with the appropriate units. Part C What is the instantaneous emf Epsilon when i = -I? Express your answer with the appropriate units.

Explanation / Answer

Inductive reactance

XL=2pifL =2pi*3000*(3*10-3) =56.55 ohms

Capacitive reactance

XC=1/2pi*f*C =1/2pi*3000*(510*10-9) =104.02 ohms

Phase angle

o=tan-1(XL-XC/R) =tan-1(56.55-104.02/47)=-45.4o

So Instantaneous Current

i=I*Cos(Wt-oo)

=>i=I*Cos(Wt+45.4)

a)

at i=I

Cos(Wt+45.4)=1

Wt+45.4 =0

Wt=-45.4o

so Instantaneous emf

E=EoCos(Wt) =4.7Cos(-45.4)

E=3.3 Volts

b)

at i=0 A

Cos(Wt+45.4)=0

Wt+45.4=90

Wt=44.6o

so Instantaneous emf

E=EoCos(Wt) =4.7Cos(44.6)

E=3.3465 Volts =3.35 Volts (approx)

c)

at i=-I

Cos(Wt+45.4)=-1

Wt+45.4 =180

Wt=134.6o

so Instantaneous emf

E=EoCos(Wt) =4.7Cos(134.6)

E=-3.3 Volts

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